# Calculating potential between two points + additional questions

Discussion in 'Homework Help' started by ZeroTorrent, Mar 28, 2012.

1. ### ZeroTorrent Thread Starter New Member

Mar 28, 2012
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Hello all! This is my first time posting on your lovely forums!

I am something of a complete idiot when it comes to circuit theory. I have done many university courses in advanced Quantum mechanics and math but circuit theory makes me want to curl up in the shower and cry

There seems to be a particular type of problem that stumps me and it makes me rather confused.

In problem 1 you can see a typical circuit of which i'm supposed to make a Thevinin circuit over R4. I use the node method as that method works (most of the time) and it's fairly easy. I take the bottom rectangle and set it as grounded and the other one as node e. The problem is that I am not sure what to do with E3 using the node method.

Node e: (current going in = current going out)

i_1 + i_2 = i_3 => (E1-e)/R1 + (E2-e)/R2 - e/R3 = 0 => E1/R1 +E2/R2 = e(1/R1+1/R2+1/R3) => inserting numbers => 0.4 = 0.05e => e = 8 Volts

I really wish I knew how to incorporate E3 in to my node equation. This is also where I get confused; Looking at a solution i've done everything right except I need to subtract the 6V of E3, so U_AB = A - 6 = 2 V. I thought it would add because of the polarity of E3: It flows perfectly in to node A and should this make the potential higher. There are other problems where I thought I'd get an addative potential when one should actually be subtracted from the other. Could someone sort out this problem and explain it for me?

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Can you post a circuit diagram?

Mar 28, 2012
10
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Ofc, sorry

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4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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You answer looks good to me.
Ve = 8V
Because positive side of the E3 voltage source is point toward Ve.
This means that "+" terminal of the E3 has a higher potential (higher by 6V) than negative terminal.
And this is why Ve = Va + E3 must be true.

Last edited: Mar 28, 2012
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5. ### ZeroTorrent Thread Starter New Member

Mar 28, 2012
10
0
Okay, I think I understand now. After using the node calculations I can fall back on KVL over node e. It's just an additinal step where as I do calculate using all currents don't get all voltages and have to account for them seperatly.

Thank you!